An Explicit Relationship Between Sequential Patterns and Their Concise Representations

Abstract

Mining sequential patterns in a sequence database (SDB) is an important and useful data mining task. Most existing algorithms for performing this task directly mine the set ${\mathcal{F}\mathcal{S}}$ of all frequent sequences in an SDB. However, these algorithms often exhibit poor performance on large SDBs due to the enormous search space and cardinality of ${\mathcal{F}\mathcal{S}}$. In addition, constraint-based mining algorithms relying on this approach must read an SDB again when a constraint is changed by the user. To address this issue, this paper proposes a novel approach for generating ${\mathcal{F}\mathcal{S}}$ from the two sets of frequent closed sequences $\left( {{\mathcal{F}\mathcal{C}\mathcal{S}}} \right)$ and frequent generator sequences $({\mathcal{F}\mathcal{G}\mathcal{S}})$, which are concise representations of ${\mathcal{F}\mathcal{S}}$. The proposed approach is based on a novel explicit relationship between ${\mathcal{F}\mathcal{S}}$ and these two sets. This relationship is the theoretical basis for a novel efficient algorithm named GFS-CR that directly enumerates ${\mathcal{F}\mathcal{S}}$ from ${\mathcal{F}\mathcal{C}\mathcal{S}}$ and ${\mathcal{F}\mathcal{G}\mathcal{S}}$ rather than mining them from an SDB. Experimental results show that GFS-CR outperforms state-of-the-art algorithms in terms of runtime and scalability.

Appendices

Appendices

1.1 Appendix 1: Proof of Proposition 1

• (i) “${\mathcal{F}\mathcal{S}}\left( {\sigma , \gamma } \right) \subseteq {\mathcal{F}\mathcal{S}} '\left( {\sigma ,\gamma } \right)$”: Consider any $\alpha \in {\mathcal{F}\mathcal{S}}\left( {\sigma , \gamma } \right):\gamma \;{ \sqsubseteq }\;\alpha \;{ \sqsubseteq }\;\sigma$. Without loss of generality, we can assume that $\gamma = E_{1} \to E_{2} \to \ldots \to E_{p}$, $\alpha = F_{1} \to F_{2} \to \ldots \to F_{q}$, $\sigma = E_{1}^{'} \to E_{2}^{'} \to \ldots \to E_{q}^{'}$, $F_{i} \subseteq E_{i}^{'}$, ∀i = 1, …, q and $\exists lp = \{ j_{1} ,j_{2} , \ldots ,j_{p} \}$ $\in LP(\alpha ,\gamma ) \subseteq LP(\sigma ,\gamma )$, with $1 \leqslant j_{1} < j_{2} < \ldots j_{p} \leqslant q$: $E_{k} \subseteq F_{jk} \subseteq E_{jk}^{'}$, ∀k = 1,…, p. Set $d_{i} \mathop = \limits^{\text{def}} \left\{ {\begin{array}{*{20}c} {F_{i} ,} & { if\;i \notin lp} \\ {F_{i} \backslash E_{k} , } & {if\;i = j_{k} \in lp} \\ \end{array} } \right.$, $D_{i} \mathop = \limits^{\text{def}} \left\{ {\begin{array}{*{20}c} {E^{{\prime }}_{i} , } & {if\; i \notin lp} \\ {E^{{\prime }}_{i} \backslash E_{k} , } & {if \;i = j_{k} \in lp} \\ \end{array} } \right.$, δ(lp) $\mathop = \limits^{\text{def}} d_{1} \to d_{2} \to \ldots \to d_{q}$, $E_{i}^{{ \sim }} \mathop = \limits^{\text{def}} \left\{ {\begin{array}{*{20}l} {\emptyset , \quad \quad if\;i \notin lp} \hfill \\ {E_{k} , \;if\; i = j_{k} \in lp} \hfill \\ \end{array} } \right.$, ∀i = 1, …, q, Ex(γ, lp) $\mathop = \limits^{\text{def}} E_{1}^{{ \sim }} \to E_{2}^{{ \sim }} \to \ldots \to E_{q}^{{ \sim }}$, then $d_{i} \subseteq D_{i}$ and $F_{i} = E_{i}^{{ \sim }} + d_{i}$, ∀i = 1,…, q, so $\alpha \, = \,Ex(\gamma, lp) \oplus \varDelta \left( {lp} \right)\, \in \,{\mathcal{F}\mathcal{S}} '\left( {\sigma ,\gamma } \right)$.

• (ii) “${\mathcal{F}\mathcal{S}}\left( {\sigma , \gamma } \right) \supseteq {\mathcal{F}\mathcal{S}} '\left( {\sigma ,\gamma } \right)$’’: $\forall \alpha \, \in \,{\mathcal{F}\mathcal{S}} '\left( {\sigma ,\gamma } \right)$, $\alpha = Ex(\gamma , \, lp) \, \oplus \, \varDelta \left( {lp} \right) = F_{1} \to F_{2} \to \ldots \to F_{q}$, then $F_{i} = E_{i}^{{ \sim }} + d_{i}$, $d_{i} \subseteq D_{i}$, $E_{i}^{{ \sim }} \subseteq F_{i} \subseteq E_{i}^{\sim} + D_{i} \subseteq E_{i}^{'}$, ∀i = 1,…, q. Thus, $\gamma \;{ \sqsubseteq }\;\alpha \;{ \sqsubseteq }\;\sigma$, $\alpha \, \in \,{\mathcal{F}\mathcal{S}}\left( {\sigma , \gamma } \right)$. □

1.2 Appendix 2: Proof of Theorem 1

• a. First, we prove that $\forall \alpha \, \in \,{\mathcal{F}\mathcal{S}}$, $\exists \sigma \in CloSet(\alpha )$ such that $ro\mathop = \limits^{\text{def}} \rho (\alpha ) = \rho (\sigma )$, so $\sigma \, \in \,{\mathcal{F}\mathcal{C}\mathcal{S}}$ and $ro \in RO$, then FS(ro) =[σ]. Indeed, $\beta \, \in \,FS\left( {ro} \right) \Leftrightarrow [\beta \, \in \,{\mathcal{F}\mathcal{S}}\, \wedge \,\rho (\beta )\, = \,ro] \Leftrightarrow [\beta \, \in \,{\mathcal{F}\mathcal{S}}\, \wedge \,\rho (\beta )\, = \,\rho (\sigma )\, = \,ro] \Leftrightarrow \beta \in [\sigma ]$.

• (i) “${\mathcal{F}\mathcal{S}}\, \subseteq \,\mathop {\bigcup }\nolimits_{ro\, \in \,RO} FS\left( {ro} \right)$”: $\forall \alpha \, \in \,{\mathcal{F}\mathcal{S}}$, $\exists \sigma \in CloSet( \alpha )\;{\text{ and}}\,ro\mathop = \limits^{\text{def}} \rho (\sigma )\, = \,\rho ( \alpha )$. Then $\alpha \in [\sigma ],supp(\sigma ) = supp( \alpha ) \geqslant ms$. Thus, $\sigma \, \in \,{\mathcal{F}\mathcal{C}\mathcal{S}}$, ro ∊ RO and $\alpha$ ∊ FS(ro).

• (ii) “$\mathop {\bigcup }\nolimits_{ro\, \in \,RO} FS\left( {ro} \right)\, \subseteq \,{\mathcal{F}\mathcal{S}}$”: $\forall ro\, \in \,RO$ and $\alpha$ ∊ FS(ro), then $\exists \sigma \, \in \,{\mathcal{F}\mathcal{C}\mathcal{S}}:\,\alpha { \sqsubseteq }\sigma$ and $ro = \rho (\sigma ) = \rho (\alpha )$, so $supp(\alpha ) = supp(\sigma ) \geqslant ms$ and $\alpha \, \in \,{\mathcal{F}\mathcal{S}}$. Hence, ${\mathcal{F}\mathcal{S}}\, = \,\mathop {\bigcup }\nolimits_{ro\, \in \,RO} FS\left( {ro} \right)$.

Since ${ \sim }$ is an equivalence relation, then different equivalence classes [σ] or FS(ro) are disjoint. Thus, ${\mathcal{F}\mathcal{S}}\, = \,\mathop {\bigcup }\nolimits_{ro\, \in \,RO} FS\left( {ro} \right)$.

• b. First, we prove that all different subsets ${\mathcal{F}\mathcal{S}}^{*} (\sigma_{i} ,\,\gamma_{j} )$, $\forall \sigma_{i} \in CS\left( {ro} \right)$, $\forall \gamma_{j} \in GenSet(\sigma_{i} )$, are disjoint. Indeed, assume conversely that $\exists \sigma_{m}$, $\sigma_{i} \in CS\left( {ro} \right)$, $\exists \gamma_{k} \in GenSet(\sigma_{m} )$, $\exists \gamma_{j} \in GenSet(\sigma_{i} )$: ($\sigma_{m} \ne \sigma_{i}$ or $\gamma_{k} \ne \gamma_{j}$) and $\exists \alpha \, \in \,{\mathcal{F}\mathcal{S}}^{*} (\sigma_{m} ,\,\gamma_{k} )\, \cap \,{\mathcal{F}\mathcal{S}}^{*} (\sigma_{i} ,\,\gamma_{j} )( \ne \emptyset )$. Consider the following two cases.

• (i) If $\sigma_{m} \ne \sigma_{i}$, then without loss of generality, we can assume that m < i and $\exists \gamma \in GenSet(\alpha )$ (because $GenSet(\alpha ) \, \ne \emptyset )$. We have $\gamma \, \in \,\mathcal{G}\mathcal{S}$, $\gamma \;{ \sqsubseteq }\;\alpha \;{ \sqsubseteq }\;\sigma_{i}$, $\rho (\gamma ) = \rho (\alpha ) = \rho (\sigma_{i} )$, because $\alpha \, \in \,{\mathcal{F}\mathcal{S}}^{*} (\sigma_{i} ,\,\gamma_{j} )$. Then $\gamma \in GenSet(\sigma_{i} )$, $\alpha \in {\mathcal{F}\mathcal{S}}(\sigma_{i} ,\gamma )$ and $\alpha \in {\mathcal{F}\mathcal{S}}^{*} (\sigma_{i} ,\gamma )$. Moreover, since $\alpha \, \in \,{\mathcal{F}\mathcal{S}}^{*} (\sigma_{m} ,\,\gamma_{k} )\, \subseteq \,{\mathcal{F}\mathcal{S}}(\sigma_{m} ,\,\gamma_{k} )$, then $\alpha { \sqsubseteq }\sigma_{m}$ with m < i. This contradicts the condition not(DCondC($\alpha$, $\sigma_{i}$, CS(ρ($\sigma_{i}$)))) in ${\mathcal{F}\mathcal{S}}^{*} (\sigma_{i} ,\,\gamma )$.

• (ii) Otherwise, if $\sigma_{m} \equiv \sigma_{i}$ and $\gamma_{k} \ne \gamma_{j}$, then without loss of generality, we can assume that k < j. Since $\alpha \, \in \,{\mathcal{F}\mathcal{S}}^{*} (\sigma_{i} ,\,\gamma_{k} ) \subseteq {\mathcal{F}\mathcal{S}}(\sigma_{i} ,\,\gamma_{k} )$, then $\alpha \;{ \sqsupseteq }\;\gamma_{k}$, with k < j. This also contradicts the condition not $(DCondG(\alpha ,\sigma_{i} ,\gamma_{j} ))$ in ${\mathcal{F}\mathcal{S}}^{*} (\sigma_{i} ,\,\gamma_{j} )$.

  Finally, we prove that $FS\left( {ro} \right) = \mathop \sum \nolimits_{{\sigma_{i} \in CS\left( {ro} \right)}} \mathop \sum \nolimits_{{\gamma_{j} \in GenSet\left( {\sigma_{i} } \right)}} {\mathcal{F}\mathcal{S}}^{*} \left( {\sigma_{i} , \gamma_{j} } \right)$, $\forall ro \in RO$.

• (iii) “⊆”: $\forall \alpha \in FS\left( {ro} \right)$, then $\alpha \in {\mathcal{F}\mathcal{S}}$ and ρ(α) = ro. Since $CloSet\left( \alpha \right) \ne \emptyset$ and $GenSet\left( \alpha \right) \ne \emptyset$, [$\exists \sigma_{i} \in CloSet(\alpha )$, $\exists \gamma_{j} \in GenSet(\alpha ):\sigma_{i} \in {\mathcal{C}\mathcal{S}},\gamma_{j} \in {\mathcal{G}\mathcal{S}}$, $\gamma_{j} \; { \sqsubseteq }\;\alpha \;{ \sqsubseteq }\;\sigma_{i}$, $\rho (\sigma_{i} ) = \rho (\gamma_{j} ) = \rho \left( \alpha \right)]^{{^{\left( * \right)} }}$, so $\gamma_{j} \in GenSet(\sigma_{i} )$, $supp(\sigma_{i} ) = supp(\gamma_{j} ) = supp\left( \alpha \right) \geqslant ms$, $\sigma_{i} \in {\mathcal{F}\mathcal{C}\mathcal{S}}$ and $\sigma_{i} \in CS\left( {ro} \right)$. Hence, $\alpha \in {\mathcal{F}\mathcal{S}}(\sigma_{i} ,\gamma_{j} )$. Without loss of generality, we can select $\sigma_{i}$ and $\gamma_{j}$ that are, respectively, the first closed and generator sequences satisfying the condition^(*). Thus, $FS\left( {ro} \right) \subseteq \mathop \sum \nolimits_{{\sigma_{i} \in CS\left( {ro} \right)}} \mathop \sum \nolimits_{{\gamma_{j} \in GenSet\left( {\sigma_{i} } \right)}} {\mathcal{F}\mathcal{S}}^{*} \left( {\sigma_{i} ,\gamma_{j} } \right)$.

• (iv) “⊇”: $\forall \sigma_{i} \in CS\left( {ro} \right)$, $\forall \gamma_{j} \in GenSet(\sigma_{i} )$, $\forall \alpha \in {\mathcal{F}\mathcal{S}}^{*} (\sigma_{i} ,\gamma_{j} )$, we have $\sigma_{i} \in {\mathcal{F}\mathcal{C}\mathcal{S}}$, $\alpha { \sqsubseteq }\sigma_{i}$, $\rho \left( \alpha \right) \, = \rho (\sigma_{i} ) = ro$, so $supp\left( \alpha \right) = supp(\sigma_{i} ) \geqslant ms$, i.e. $\alpha \in FS\left( {ro} \right)$ . Thus, ${\mathcal{F}\mathcal{S}}^{*} \left( {\sigma_{i} ,\gamma_{j} } \right) \subseteq FS\left( {ro} \right)\,{\text{and}}\,FS\left( {ro} \right) \supseteq \sum\nolimits_{{\sigma_{i} \in CS\left( {ro} \right)}} {\sum\nolimits_{{\gamma_{j} \in GenSet\left( {\sigma_{i} } \right)}} {{\mathcal{F}\mathcal{S}}^{*} \left( {\sigma_{i} ,\gamma_{j} } \right)} }$. □

1.3 Appendix 3: Proof of Theorem 2

• (a) It is clear that ${\mathcal{F}\mathcal{S}}^{**} \left( {\sigma ,\gamma } \right) \subseteq { \mathcal{F}\mathcal{S}}^{{\prime }} \left( {\sigma , \gamma } \right)$. In the proof of the above three pruning cases, $\forall \alpha \in {\mathcal{F}\mathcal{S}}^{{\prime }} \left( {\sigma , \gamma } \right)$, if ∃i = 1, 2, 3: $DCond_{i}$ is true, then α is a duplicate of some previously generated sequences. Thus, ${\mathcal{F}\mathcal{S}} '\left( {\sigma , \gamma } \right) \subseteq {\mathcal{F}\mathcal{S}}^{**} \left( {\sigma ,\gamma } \right)$ and ${\mathcal{F}\mathcal{S}} '\left( {\sigma , \gamma } \right) = {\mathcal{S}}^{**} \left( {\sigma ,\gamma } \right)$.

• (b) We will prove the first assertion by contradiction. Assume conversely that there are two sequences $\alpha = Ex(\gamma ,lp_{m} )$ ⊕ $\varDelta (lp_{m} )$ and $\beta = Ex(\gamma ,lp_{n} ) \oplus$, $\varDelta (lp_{n} )$ (in ${\mathcal{F}\mathcal{S}}^{**} \left( {\sigma ,\gamma } \right)$) according to two different position lists in LP(σ, γ): $lp_{m} \ne lp_{n}$, but α ≡ β, i.e., there exist two $lp_{m} = \{ j_{1} ,j_{2} , \ldots ,j_{p} \}$, $lp_{n} = \{ i_{1} ,i_{2} , \ldots ,i_{p} \} \in LP\left( {\sigma ,\gamma } \right)$ with m < n, $1 \leqslant j_{1} < j_{2} < \ldots j_{p} \geqslant q$, $1 \leqslant i_{1} < i_{2} < \ldots < i_{p} \leqslant q$, $d(lp_{m} )_{i} \mathop = \limits^{\text{def}} \left\{ {\begin{array}{*{20}l} {E^{{\prime }}_{i} , \quad \quad if\, i \notin lp_{m} } \hfill \\ {E^{{\prime }}_{i} \backslash E_{k} , \, if\, i = j_{k} \in lp_{m} } \hfill \\ \end{array} } \right.$, $d_{i} \mathop = \limits^{\text{def}} d(lp_{n} )_{i} \mathop = \limits^{\text{def}} \left\{ {\begin{array}{*{20}l} {E_{i}^{'} ,\quad \quad if \,i \notin lp_{n} } \hfill \\ {E^{{\prime }}_{i} \backslash E_{k} , \, if\, i = i_{k} \in lp_{n} } \hfill \\ \end{array} } \right.$, $F_{i} \subseteq d\left( {lp_{m} } \right)_{i}$, $F^{{\prime }}_{i} \subseteq d_{i}$, ∀i = 1,…, q, and two corresponding sequences $\alpha = F_{1} \to F_{2} \to \ldots \to F_{q}$ and $\beta = F_{1}^{'} \to F_{2}^{'} \to \ldots \to F^{{\prime }}_{q}$ that belong to ${\mathcal{F}\mathcal{S}}^{**} \left( {\sigma ,\gamma } \right)$ such that α ≡ β and $DCond_{k} (lp_{m} )$, $DCond_{k} (lp_{n} )$ are false, ∀k = 1, 2, 3.

After deleting all empty itemsets in α and β, we obtain $\alpha = F_{{u_{1} }} \to F_{{u_{2} }} \to \ldots \to F_{{u_{N} }}$, $\beta = F^{{\prime }}_{{v_{1} }} \to F^{{\prime }}_{{v_{2} }} \to \ldots \to F^{{\prime }}_{{v_{N} }}$ with N = size(α) = size(β), $1 \leqslant u_{1} < u_{2} < \ldots < u_{N} \leqslant q$, $1 \leqslant v_{1} < v_{2} < \ldots < v_{N} \leqslant q$ and $F_{{u_{i} }} = F^{{\prime }}_{{v_{i} }} \ne \emptyset ,\forall i = 1, \ldots ,N$ (because α ≡ β). We set $i_{0} = j_{0} = u_{0} = v_{0} = k_{0} \equiv \, 0$, $k_{p + 1} = N$ and $1 \leqslant k_{1} < k_{2} < \ldots < k_{p} \leqslant N$ : $j_{r} = u_{{k_{r} }}$ (so $F_{{u_{{k_{r} }} }} = F_{{j_{r} }} \supseteq E_{r} \subseteq F^{{\prime }}_{{j_{r} }}$), ∀r = 1, …, p.

For ∀r = 1, …, p + 1, $\forall k = \left( {k_{r - 1} + 1} \right), \ldots , k_{r}$, then $u_{{k_{r - 1} }} = j_{r - 1} < u_{k} \leqslant j_{r} \leqslant i_{r}$ and we prove that if {($v_{h} = u_{h}$, ∀h = 0, …, k − 1), $\left( {v_{k - 1} < i_{r} } \right)$ and $(\forall r^{{\prime }} = \, 0, \, ..,r - 1,j_{{r^{{\prime }} }} = i_{{r^{{\prime }} }} = u_{{k_{{r^{{\prime }} }} }} = v_{{k_{{r^{{\prime }} }} }} )\}$, (H(r, k)), then {$v_{k} = u_{k}$, and if (r ⩽ p) then [($v_{k} < i_{r}$, if $k < k_{r}$) and $\left( {j_{r} = i_{r} = u_{{k_{r} }} = v_{{k_{r} }} } \right)$]}, (C(r, k)).

Since $i_{0} = j_{0} = u_{0} = v_{0} = k_{0} = 0$ and $i_{0} < i_{1}$, then the hypothesis H(1, 1) is always true. For ∀r = 1, …, p + 1, $\forall k = \, (k_{r - 1} + 1), ..,k_{r}$, assume that the hypothesis H(r, k) is true.

• (i) First, we consider any k such that [$k_{r - 1} + 1 \leqslant k < k_{r}$, for r ⩽ p] or [$k_{p} + 1 \leqslant k \leqslant N$, with r = p + 1]. If r ⩽ p, then $i_{r - 1} = j_{r - 1} = u_{{k_{r - 1} }} < u_{k} < j_{r} = u_{{k_{r} }}$ and $v_{k} \leqslant i_{r}$, because $F_{{u_{k} }} = F^{{\prime }}_{{v_{k} }} \ne \emptyset$ and $v_{k - 1} < i_{r}$. Consider the following three cases.

  If ($r \leqslant p$ and $v_{k} = i_{r}$): then $j_{r - 1} < u_{k} < j_{r} \leqslant v_{k} = i_{r}$, $E^{{\prime }}_{{u_{k} }} \supseteq F_{{u_{k} }} = F^{{\prime }}_{{v_{k} }} = F^{{\prime }}_{{i_{r} }} \supseteq E_{r}$, so $u_{k} \in lp_{pre} \mathop = \limits^{\text{def}} \{ j_{1} , \ldots ,j_{r - 1} , u_{k} ,j_{r + 1} , \ldots ,j_{p} \}$ (and $lp_{pre} \mathop = \limits^{\text{def}} \{ u_{k} ,j_{2} , \ldots ,j_{p} \}$ if r = 1), $lp_{pre} \prec lp_{m}$, FI(pre, m) = r, $lp_{pre} \left[ r \right] \, = u_{k}$ and the $u_{k}^{th}$ itemset $\left( {F_{{u_{k} }} } \right)$ contains $E_{r}$, so $DCond_{2} (lp_{pre} ,lp_{\varvec{m}} ,u_{k} )$ is true. Therefore, $v_{k} < i_{r}$.

  If ($u_{k} < v_{k}$ and ($v_{k} < i_{r}$, if $r \leqslant p$)): then $i_{r - 1} = j_{r - 1} = u_{{k_{r - 1} }} \leqslant u_{k - 1} = v_{k - 1} < u_{k} < v_{k}$ and ($v_{k} < i_{r}$, with $r \leqslant p$), $v_{k} \notin lp_{n}$, $u_{k} \notin lp_{n}$, $d\left( {lp_{n} } \right)_{{v_{k} }} = E^{{\prime }}_{{v_{k} }}$, $d\left( {lp_{n} } \right)_{{u_{k} }} = E^{{\prime }}_{{u_{k} }}$, $F^{{\prime }}_{{v_{k - 1} }} \ne \emptyset$, $F^{{\prime }}_{r'} = \emptyset$, $\forall r^{{\prime }} = (v_{k - 1} + 1), \ldots ,(v_{k} - 1)$, $v_{k - 1} + 1 \le u_{k} \le v_{k} - 1$. Moreover, for the non-empty itemset $its_{{t_{v_{k}} }}$ of $\beta$ such that $its_{{t_{v_{k}} }} = F^{\prime }_{{v_{k} }} = F_{{u_{k} }} \subseteq E^{\prime }_{{v_{k} }} \cap E^{\prime }_{{u_{k} }} = d\left( {lp_{n} } \right)_{{v_{k} }} \cap d\left( {lp_{n} } \right)_{{u_{k} }}$, so $DCond_{1} (lp_{n} ,v_{k} ,u_{k} )$ is true.

  If $v_{k} < u_{k}$, then $j_{r - 1} = u_{{k_{r - 1} }} \le u_{k - 1} = v_{k - 1} < v_{k} < u_{k}$ and ($u_{k} < j_{r}$, with $r \leqslant p$). Similarly, $u_{k} \notin lp_{m}$, $v_{k} \notin lp_{m}$, $d\left( {lp_{m} } \right)_{{v_{k} }} = E^{{\prime }}_{{v_{k} }}$, $d\left( {lp_{m} } \right)_{{u_{k} }} = E^{{\prime }}_{{u_{k} }}$, $F_{{u_{k - 1} }} \ne \emptyset$, $F_{{r^{{\prime }} }} = \emptyset$, $\forall r^{{\prime }} = (u_{k - 1} + 1), \ldots ,(u_{k} - 1)$, $u_{k - 1} + 1 \leqslant v_{k} \leqslant u_{k} - 1$. Moreover, for the non-empty itemset $its_{{u_{k} }}$ of $\alpha$ such that $its_{{u_{k} }} = F_{{u_{k} }} = F_{{v_{k} }}^{{\prime }} \subseteq E_{{u_{k} }}^{{\prime }} \cap E_{{v_{k} }}^{{\prime }} = d\left( {lp_{m} } \right)_{{u_{k} }} \cap d\left( {lp_{m} } \right)_{{v_{k} }}$, so $DCond_{1} (lp_{m} ,u_{k} ,v_{k} )$ is true.

  Hence, $u_{k} = v_{k}$ and ($v_{k} < j_{r} \leqslant i_{r}$, with $r \leqslant p$). $\left( {C\left( {r, \, k} \right)^{(1)} } \right)$.

• (ii) Second, if $r \leqslant p$, we consider $k = k_{r}$. Then $i_{r - 1} = j_{r - 1} = u_{{k_{r - 1} }} < u_{k} = u_{{k_{r} }} = j_{r} \le i_{r}$ Since $F_{{u_{k} }} = F^{{\prime }}_{{v_{k} }} \ne \emptyset$ and $v_{k - 1} = v_{{k_{r} - 1}} < i_{r}$, then $v_{{k_{r} }} \le i_{r}$. Consider the following cases.

If $v_{{k_{r} }} < u_{{k_{r} }} = j_{r} \le i_{r}$, then $i_{r - 1} = j_{r - 1} = u_{{k_{r - 1} }} = v_{{k_{r - 1} }} < v_{{k_{r} }} < u_{{k_{r} }} = j_{r} \le i_{r}$, $E^{{\prime }}_{{v_{{k_{r} }} }} \supseteq F^{{\prime }}_{{v_{{k_{r} }} }} = F_{{u_{{k_{r} }} }} = F_{{j_{r} }} \supseteq E_{r}$. Thus, $v_{{k_{r} }} \in lp_{pre} \mathop = \limits^{\text{def}} \{ i_{1} , \ldots ,i_{r - 1} , v_{{k_{r} }} ,i_{r + 1} , \ldots ,i_{p} \}$ (and $lp_{pre} \mathop = \limits^{\text{def}} \{ v_{{k_{1} }} ,i_{2} , \ldots ,i_{p} \}$, if r = 1), $lp_{pre} \prec lp_{n}$, $FI\left( {pre,n} \right) \, = r,lp_{pre} \left[ r \right] \, = v_{{k_{r} }}$ and with the itemset $F^{{\prime }}_{{v_{{k_{r} }} }}$ of $\beta$ such that $F^{{\prime }}_{{v_{{k_{r} }} }} \supseteq E_{r}$, so $DCond_{2} (lp_{pre} ,lp_{n} ,v_{{k_{r} }} )$ is true (or $DCond_{3} (lp_{pre} \mathop = \limits^{\text{def}} \{ j_{1} , \ldots ,j_{r - 1} , v_{{k_{r} }} ,j_{r + 1} , \ldots ,j_{p} \} ,lp_{m} ,u_{{k_{r} }} )$ is true).

Therefore, $j_{r} = u_{{k_{r} }} \leqslant v_{{k_{r} }} \leqslant i_{r}$. But the case $u_{{k_{r} }} < v_{{k_{r} }} \le i_{r}$ cannot occur.

Indeed, if $j_{r} = u_{{k_{r} }} < v_{{k_{r} }} = i_{r}$, then $i_{r - 1} = j_{r - 1} = u_{{k_{r - 1} }} = v_{{k_{r - 1} }} \leqslant u_{{k_{r} - 1}} < u_{{k_{r} }} = j_{r} < v_{{k_{r} }} = i_{r}$. Thus, $F^{{\prime }}_{{r^{{\prime }} }} = \emptyset$, $\forall r^{{\prime }} = u_{{k_{r} }} , \ldots ,(v_{{k_{r} }} - 1)$, $E^{{\prime }}_{{u_{{k_{r} }} }} \supseteq F_{{u_{{k_{r} }} }} = F_{{j_{r} }} \supseteq E_{r}$, $u_{{k_{r} }} \in lp_{pre} \mathop = \limits^{\text{def}} \{ i_{1} , \ldots ,i_{r - 1} , u_{{k_{r} }} ,i_{r + 1} , \ldots ,i_{p} \}$ (and $lp_{pre} \mathop = \limits^{\text{def}} \{ u_{{k_{1} }} ,i_{2} , \ldots ,i_{p} \}$, if r = 1), $lp_{pre} \prec lp_{n}$, FI(pre, n) = r, $lp_{pre} \left[ r \right] \, = u_{{k_{r} }}$, $lp_{n} \left[ r \right] \, = i_{r} = v_{{k_{r} }}$. Since $its_{{v_{{k_{r} }} }} = F^{{\prime }}_{{v_{{k_{r} }} }} \backslash E_{r} \subseteq d(lp_{n} )_{{v_{{k_{r} }} }} = E^{{\prime }}_{{v_{{k_{r} }} }} \backslash E_{r}$ and $F_{{u_{{k_{r} }} }} \backslash E_{r} \subseteq d(lp_{pre} )_{{u_{{k_{r} }} }} = E^{{\prime }}_{{u_{{k_{r} }} }} \backslash E_{r}$, then for itemset $its_{{v_{{k_{r} }} }}$ of $\beta$ such that $its_{{v_{{k_{r} }} }} = F^{{\prime }}_{{v_{{k_{r} }} }} \backslash E_{r} = F_{{u_{{k_{r} }} }} \backslash E_{r} \subseteq d(lp_{n} )_{{v_{{k_{r} }} }} \cap d(lp_{pre} )_{{u_{{k_{r} }} }}$. Therefore, $DCond_{3} (lp_{pre} ,lp_{n} ,v_{{k_{r} }} )$ is true.

If $j_{r} = u_{{k_{r} }} < v_{{k_{r} }} < i_{r}$, then $i_{r - 1} = j_{r - 1} = u_{{k_{r - 1} }} = v_{{k_{r - 1} }} \leqslant v_{{k_{r} - 1}} = u_{{k_{r} - 1}} < u_{{k_{r} }} = j_{r} < v_{{k_{r} }} < i_{r}$. Hence, $F^{{\prime }}_{{r^{{\prime }} }} = \emptyset$, $\forall r^{{\prime }} = \, (v_{{k_{r} - 1}} + 1), \, .., \, (v_{{k_{r} }} - 1)$, $v_{{k_{r} - 1}} + 1 \leqslant u_{{k_{r} }} \leqslant v_{{k_{r} }} - 1,u_{{k_{r} }}$ and $v_{{k_{r} }} \notin lp_{n} ,d\left( {lp_{n} } \right)_{{v_{{k_{r} }} }} = E^{{\prime }}_{{v_{{k_{r} }} }} ,F_{{u_{{k_{r} }} }} \subseteq d\left( {lp_{n} } \right)_{{u_{{k_{r} }} }} = E^{{\prime }}_{{u_{{k_{r} }} }}$. Thus, for the non-empty itemset $its_{{v_{{k_{r} }} }}$ of $\beta$ such that $its_{{v_{{k_{r} }} }} = F^{{\prime }}_{{v_{{k_{r} }} }} = F_{{u_{{k_{r} }} }} \subseteq d\left( {lp_{n} } \right)_{{v_{{k_{r} }} }} \cap d\left( {lp_{n} } \right)_{{u_{{k_{r} }} }}$ , the condition $DCond_{1}$($lp_{n}$, $v_{{k_{r} }}$, $u_{{k_{r} }}$) is true.

Thus, $v_{{k_{r} }} = u_{{k_{r} }} = j_{r} \leqslant i_{r}$. However, if $v_{{k_{r} }} = u_{{k_{r} }} = j_{r} < i_{r}$, then $i_{r - 1} = j_{r - 1} = u_{{k_{r - 1} }} = v_{{k_{r - 1} }} < v_{{k_{r} }} = u_{{k_{r} }} = j_{r} < i_{r}$, $E^{{\prime }}_{{j_{r} }} \supseteq F^{{\prime }}_{{j_{r} }} = F^{'}_{{v_{{k_{r} }} }} = F_{{u_{{k_{r} }} }} \supseteq E_{r}$. Thus, $j_{r} \in lp_{pre} \mathop = \limits^{\text{def}} \{ i_{1} , \ldots ,i_{r - 1} , j_{r} ,i_{r + 1} , \ldots ,i_{p} \}$ (and $lp_{pre} \mathop = \limits^{\text{def}} \{ j_{1} ,i_{2} , \ldots ,i_{p} \}$, if r = 1), $lp_{pre} \prec lp_{n}$, FI(pre, n) = r, $lp_{pre}$[r] = $j_{r}$ and the itemset $F^{{\prime }}_{{j_{r} }}$ of $\beta$ contains $E_{r}$, so $DCond_{2}$($lp_{pre}$, $lp_{n}$, $j_{r}$) is true.

Hence, $v_{{k_{r} }} = u_{{k_{r} }} = j_{r} = i_{r}$. (C(r, k)⁽²⁾).

From (C(r, k) ⁽¹⁾) − (C(r, k) ⁽²⁾), we have (C(r, k)).

Finally, under the hypothesis α ≡ β, then ∀r = 1, …, p + 1, ∀k = $(k_{r - 1} + 1), \ldots , k_{r}$, we always have ($v_{k} = u_{k}$), ($j_{r} = i_{r} = u_{{k_{r} }} = v_{{k_{r} }}$, if $r \leqslant p$). Therefore,

$$ (\forall k^{{\prime }} = 1, \ldots ,N,v_{{k^{{\prime }} }} = u_{{k^{{\prime }} }} )^{(*)} \,{\text{and}}\, (\forall r = 1, \ldots ,p,j_{r} = i_{r} )^{(**)} . $$

If $lp_{m} = lp_{n}$, then $j_{r} = i_{r}$, ∀r = 1,.., p. Since α and β are generated from two different position lists in ${\mathcal{F}\mathcal{S}}^{**} \left( {\sigma ,\gamma } \right)$, so $\exists i_{0} = 1, \ldots ,N$ such that $u_{{i_{0} }} \ne v_{{i_{0} }}$. This contradicts ^(*).

If $lp_{m} \ne lp_{n}$, i.e. $\exists m_{0} = 1, \ldots ,p$ such that $i_{{m_{0} }} \ne j_{{m_{0} }}$. This also contradicts ^(**). In other words, the hypothesis α ≡ β always leads to a contradiction.

Thus, α ≠ β, i.e., all sequences in ${\mathcal{F}\mathcal{S}}^{**} \left( {\sigma ,\gamma } \right)$ are distinctly generated.

From Proposition 1 and Theorem 1.a, we have ${\mathcal{F}\mathcal{S}}\left( {\sigma ,\gamma } \right) = {\mathcal{F}\mathcal{S}}^{{\prime }} \left( {\sigma ,\gamma } \right) = {\mathcal{F}\mathcal{S}}^{**} \left( {\sigma ,\gamma } \right)$ and the remaining assertions are deduced from Theorem 1. □